X-ray of wavelength `lamda=2` Å is emitted from the metal target. The potential difference applied across the cathode and the metal target is

Options

A. 5525 V
B. 320 V
C. 6200 V
D. 3250 V

Answer

C

Explanation

From conservation of energy the electron kinetic energy equals the maximum photon energy (we neglect the work function `phi`, because it is normally, so small compared to `eV_(0)`)
`eV_(0)=hv_("max") or eV_(0)=(hc)/(lamda_("min"))`
`therefore v_(0)=(hc)/(elamda_("min")) or V_(0)=(12400xx10^(-10))/(10^(-11))=124kV`
Hence, accelerating voltage for electrons in X-ray machine should be less than 124 kV.